Cho $\int\limits^{\frac{\pi}{2}}_{a-2}\cos\left(x\right)^{\sin x}dx=a$ và $\int\limits^{\pi}_{a-2}\sqrt{\tan\left(x\right)}dx=2a-4$ ( với a là số nguyên dương ). Khi này tính: \(\int\limits^{a 2}

$\intπ20cos2xsin2xdx=12\intπ20cos2x(1-cos2x)dx=12\intπ20[cos2x-1+cos4x2]dx=14\intπ20(2cos2x-cos4x-1)dx=14[sin2x-sin4x4-x]π20=-14.π2=-π8\int0π2cos2xsin2xdx=12\int0π2cos2x(1-cos2x)dx=12\int0π2[cos2x-1+cos4x2]dx=14\int0π2(2cos2x-cos4x-1)dx=14[sin2x-sin4x4-x]0π2=-14.π2=-π8$

b)

Ta có: Xét 2x – 2-x ≥ 0 ⇔ x ≥ 0.

Ta tách thành tổng của hai tích phân:

$\int1-1|2x-2-x|dx=-\int0-1(2x-2-x)dx+\int10(2x-2-x)dx=-(2xln2+2-xln2)∣∣0-1+(2xln2+2-xln2)∣∣10=1ln2\int-11|2x-2-x|dx=-\int-10(2x-2-x)dx+\int01(2x-2-x)dx=-(2xln2+2-xln2)|-10+(2xln2+2-xln2)|01=1ln2$

c)

$\int21(x+1)(x+2)(x+3)x2dx=\int21x3+6x2+11x+6x2dx=\int21(x+6+11x+6x2)dx=[x22+6x+11ln|x|-6x]∣∣21=(2+12+11ln2-3)-(12+6-6)=212+11ln2\int12(x+1)(x+2)(x+3)x2dx=\int12x3+6x2+11x+6x2dx=\int12(x+6+11x+6x2)dx=[x22+6x+11ln|x|-6x]|12=(2+12+11ln2-3)-(12+6-6)=212+11ln2$

d)

$\int201x2-2x-3dx=\int201(x+1)(x-3)dx=14\int20(1x-3-1x+1)dx=14[ln|x-3|-ln|x+1|]∣∣20=14[1-ln2-ln3]=14(1-ln6)\int021x2-2x-3dx=\int021(x+1)(x-3)dx=14\int02(1x-3-1x+1)dx=14[ln|x-3|-ln|x+1|]|02=14[1-ln2-ln3]=14(1-ln6)$

e)

$\intπ20(sinx+cosx)2dx=\intπ20(1+sin2x)dx=[x-cos2x2]∣∣π20=π2+1\int0π2(sinx+cosx)2dx=\int0π2(1+sin2x)dx=[x-cos2x2]|0π2=π2+1$

g)

$I=\intπ0(x+sinx)2dx\intπ0(x2+2xsinx+sin2x)dx=[x33]∣∣π0+2\intπ0xsinxdx+12\intπ0(1-cos2x)dxI=\int0π(x+sinx)2dx\int0π(x2+2xsinx+sin2x)dx=[x33]|0π+2\int0πxsinxdx+12\int0π(1-cos2x)dx$

Tính : $J=\intπ0xsinxdxJ=\int0πxsinxdx$

Đặt u = x ⇒ u’ = 1 và v’ = sinx ⇒ v = -cos x

Suy ra:

$J=[-xcosx]∣∣π0+\intπ0cosxdx=π+[sinx]∣∣π0=πJ=[-xcosx]|0π+\int0πcosxdx=π+[sinx]|0π=π$

Do đó:

$I=π33+2π+12[x-sin2x2]∣∣π30=π33+2π+π2=2π3+15π6$

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SG

Sách Giáo Khoa

14 tháng 4 2017

Tính : a) $\int\limits^2_{-1}\left(5x^2-x+e^{0,5x}\right)dx$ b) $\int\limits^2_{0,5}\left(2\sqrt{x}+\dfrac{3}{x^2}+\cos x\right)dx$ c) $\int\limits^2_1\dfrac{dx}{\sqrt{2x+3}}$ (đặt $t=\sqrt{2x+3}$ ) d) $\int\limits^2_1\sqrt[3]{3x^3+4}x^2dx$ (đặt $t=\sqrt[3]{3x^3+4}$ ) e) $\int\limits^2_{-2}\left(x-2\right)\left|x\right|dx$ g) $\int\limits^0_1x\cos xdx$ h) \(\int\limits^{\dfrac{\pi}{2}}_{\dfrac{\pi}{6}}\dfrac{1+\sin2x+\cos2x}{\sin x+\cos...

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#Hóa học lớp 12

0

SG

Sách Giáo Khoa

12 tháng 4 2017

Tính các tích phân sau :

a) $\int\limits^{\dfrac{\pi}{4}}_0\cos2x.\cos^2xdx$

b) $\int\limits^1_{\dfrac{1}{2}}\dfrac{e^x}{e^{2x}-1}dx$

c) $\int\limits^1_0\dfrac{x+2}{x^2+2x+1}\ln\left(x+1\right)dx$

d) $\int\limits^{\dfrac{\pi}{4}}_0\dfrac{x\sin x+\left(x+1\right)\cos x}{x\sin x+\cos x}dx$

#Hóa học lớp 12

3

AH

Akai Haruma

Giáo viên

8 tháng 7 2017

a)

Ta có $A=\int ^{\frac{\pi}{4}}_{0}\cos 2x\cos^2xdx=\frac{1}{4}\int ^{\frac{\pi}{4}}_{0}\cos 2x(\cos 2x+1)d(2x)$

$\Leftrightarrow A=\frac{1}{4}\int ^{\frac{\pi}{2}}_{0}\cos x(\cos x+1)dx=\frac{1}{4}\int ^{\frac{\pi}{2}}_{0}\cos xdx+\frac{1}{8}\int ^{\frac{\pi}{2}}_{0}(\cos 2x+1)dx$

$\Leftrightarrow A=\frac{1}{4}\left.\begin{matrix} \frac{\pi}{2}\\ 0\end{matrix}\right|\sin x+\frac{1}{16}\left.\begin{matrix} \frac{\pi}{2}\\ 0\end{matrix}\right|\sin 2x+\frac{1}{8}\left.\begin{matrix} \frac{\pi}{2}\\ 0\end{matrix}\right|x=\frac{1}{4}+\frac{\pi}{16}$

b)

$B=\int ^{1}_{\frac{1}{2}}\frac{e^x}{e^{2x}-1}dx=\frac{1}{2}\int ^{1}_{\frac{1}{2}}\left ( \frac{1}{e^x-1}-\frac{1}{e^x+1} \right )d(e^x)$

$\Leftrightarrow B=\frac{1}{2}\left.\begin{matrix} 1\\ \frac{1}{2}\end{matrix}\right|\left | \frac{e^x-1}{e^x+1} \right |\approx 0.317$

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AH

Akai Haruma

Giáo viên

8 tháng 7 2017

c)

Có $C=\int ^{1}_{0}\frac{(x+2)\ln(x+1)}{(x+1)^2}d(x+1)$.

Đặt $x+1=t$

$\Rightarrow C=\int ^{2}_{1}\frac{(t+1)\ln t}{t^2}dt=\int ^{2}_{1}\frac{\ln t}{t}dt+\int ^{2}_{1}\frac{\ln t}{t^2}dt$

$=\int ^{2}_{1}\ln td(\ln t)+\int ^{2}_{1}\frac{\ln t}{t^2}dt=\frac{\ln ^22}{2}+\int ^{2}_{1}\frac{\ln t}{t^2}dt$

Đặt $\left\{\begin{matrix} u=\ln t\\ dv=\frac{dt}{t^2}\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{dt}{t}\\ v=\frac{-1}{t}\end{matrix}\right.\Rightarrow \int ^{2}_{1}\frac{\ln t}{t^2}dt=\left.\begin{matrix} 2\\ 1\end{matrix}\right|-\frac{\ln t+1}{t}=\frac{1}{2}-\frac{\ln 2 }{2}$

$\Rightarrow C=\frac{1}{2}-\frac{\ln 2}{2}+\frac{\ln ^22}{2}$

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SG

Sách Giáo Khoa

1 tháng 4 2017

Tính các tích phân sau...

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#Toán lớp 12

1

LN

Linh Nguyễn

1 tháng 4 2017

a) $\int_{\frac{-1}{2}}^{\frac{1}{2}}\sqrt[3]{ (1-x)^{2}}dx$ = $-\int_{\frac{-1}{2}}^{\frac{1}{2}}(1-x)^{\frac{2}{3}}d(1-x)=-\frac{3}{5}(1-x)^{\frac{5}{3}}|_{\frac{-1}{2}}^{\frac{1}{2}}$

= $-\frac{3}{5}\left [ \frac{1}{2\sqrt[3]{4}}-\frac{3\sqrt[3]{9}}{2\sqrt[3]{4}} \right ]=\frac{3}{10\sqrt[3]{4}}(3\sqrt[3]{9}-1)$

b) $\int_{0}^{\frac{\pi}{2}}sin(\frac{\pi}{4}-x)dx$ = $-\int_{0}^{\frac{\pi}{2}}sin(\frac{\pi}{4}-x)d(\frac{\pi}{4}-x)$ = $cos(\frac{\pi}{4}-x)|_{0}^{\frac{\pi}{2}}$

= $cos(\frac{\pi}{4}-\frac{\pi}{2})-cos\frac{\pi}{4}=0$

c) $\int_{\frac{1}{2}}^{2}\frac{1}{x(x+1)}dx$ = $\int_{\frac{1}{2}}^{2}(\frac{1}{x}-\frac{1}{x+1})dx =ln\left | \frac{x}{x+1} \right ||_{\frac{1}{2}}^{2}=ln2$

d) $\int_{0}^{2}x(x+1)^{2}dx$ = $\int_{0}^{2}(x^{3}+2x^{2}+x)dx=(\frac{x^{4}}{4}+\frac{2}{3}x^{3}+\frac{x^{2}}{2})|_{0}^{2}$

= $\frac{16}{4}+\frac{16}{3}+2= 11\tfrac{1}{3}$

e) $\int_{\frac{1}{2}}^{2}\frac{1-3x}{(x+1)^{2}}dx$ = $\int_{\frac{1}{2}}^{2}\frac{-3(x+1)+4}{(x+1)^{2}}dx=\int_{\frac{1}{2}}^{2}\left [ \frac{-3}{x+1}+\frac{4}{(x+1)^{2}} \right ]dx$